On the Euclidean plane, an ant travels from point $A(0,1)$ to point $B(d,1)$ for an integer $d$.

In each step, the ant at point $(x_0,y_0)$ chooses one of the lattice points $(x_1,y_1)$ which satisfy $x_1\ge 0$ and $y_1\ge 1$ and goes straight to $(x_1,y_1)$ at a constant velocity $v$. The value of $v$ depends on $y_0$ and $y_1$ as follows:

• If $y_0=y_1$, the value of $v$ equals $y_0$.
• If $y_0\ne y_1$, the value of $v$ equals $(y_1-y_0)/(\ln (y_1)-\ln (y_0))$.

The left image is one of the possible paths for $d=4$. First the ant goes from $A(0,1)$ to $P_1(1,3)$ at velocity $(3-1)/(\ln (3)-\ln (1))\approx 1.8205$. Then the required time is $\sqrt{5}/1.8205\approx 1.2283$.
From $P_1(1,3)$ to $P_2(3,3)$ the ant travels at velocity $3$ so the required time is $2/3\approx 0.6667$. From $P_2(3,3)$ to $B(4,1)$ the ant travels at velocity $(1-3)/(\ln (1)-\ln (3))\approx 1.8205$ so the required time is $\sqrt{5}/1.8205\approx 1.2283$.
Thus the total required time is $1.2283+0.6667+1.2283=3.1233$.

The right image is another path. The total required time is calculated as $0.98026+1+0.98026=2.96052$. It can be shown that this is the quickest path for $d=4$.

Let $F(d)$ be the total required time if the ant chooses the quickest path. For example, $F(4)\approx 2.960516287$.
We can verify that $F(10)\approx 4.668187834$ and $F(100)\approx 9.217221972$.

Find $F(10000)$. Give your answer rounded to nine decimal places.